Problem: In Takahashi's mind, there is always an integer sequence of length 2 \times 10^9 + 1: A = (A_{-10^9}, A_{-10^9 + 1}, ..., A_{10^9 - 1}, A_{10^9}) and an integer P.

Initially, all the elements in the sequence A in Takahashi's mind are 0, and the value of the integer P is 0.

When Takahashi eats symbols `+`, `-`, `>` and `<`, the sequence A and the integer P will change as follows:

* When he eats `+`, the value of A_P increases by 1;
* When he eats `-`, the value of A_P decreases by 1;
* When he eats `>`, the value of P increases by 1;
* When he eats `<`, the value of P decreases by 1.



Takahashi has a string S of length N. Each character in S is one of the symbols `+`, `-`, `>` and `<`. He chose a pair of integers (i, j) such that 1 \leq i \leq j \leq N and ate the symbols that are the i-th, (i+1)-th, ..., j-th characters in S, in this order. We heard that, after he finished eating, the sequence A became the same as if he had eaten all the symbols in S from first to last. How many such possible pairs (i, j) are there?

Constraints

* 1 \leq N \leq 250000
* |S| = N
* Each character in S is `+`, `-`, `>` or `<`.

Input

Input is given from Standard Input in the following format:


N
S


Output

Print the answer.

Examples

Input

5
+>+<-


Output

3


Input

5
+>+-<


Output

5


Input

48
-+><<><><><>>>+-<<>->>><<><<-+<>><+<<>+><-+->><<


Output

475

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

#define ll long long
ll base = 23333 , MOD = 20190816170251;
ll mul(ll a , ll b){a %= MOD; b %= MOD; ll p = ((long double)a * b) / MOD , ans = a * b - MOD * p; return (ans + MOD) % MOD;}
ll upd(ll x){return x + (x >> 63 & MOD);}
char str[250003]; int N; unordered_map < ll , int > now; ll val , inv , ans , add , mrk;

ll poww(ll a , ll b){
	ll tms = 1;
	while(b){
		if(b & 1) tms = mul(tms , a);
		a = mul(a , a); b >>= 1;
	}
	return tms;
}

int main(){
	scanf("%d %s" , &N , str + 1); inv = poww(base , MOD - 2);
	for(int i = N ; i ; --i)
		switch(str[i]){
		case '+': val = upd(val - MOD + 1); break;
		case '-': val = upd(val - 1); break;
		case '>': val = val * base % MOD; break;
		case '<': val = mul(val , inv);
		}
	mrk = 1;
	for(int i = N ; i ; --i){
		++now[mul(upd(-add) , mrk)];
		switch(str[i]){
		case '+': add = upd(add - MOD + 1); break;
		case '-': add = upd(add - 1); break;
		case '>': add = mul(add , base); mrk = mul(mrk , inv); break;
		case '<': add = mul(add , inv); mrk = mul(mrk , base);
		}
		ans += now[mul(upd(val - add) , mrk)];
	}
	cout << ans; return 0;
}